OA: E
1). X is a perfect square.
X can be \(121,144,169...\) and so on.
Unit digit of X can be even or odd
Statement 1 alone is not sufficient.

2). X is a palindrome.
X can be \(121,131,141,.....212,222\) and so on.
Unit digit of X can be even or odd
Statement 2 alone is not sufficient.

Combining \((1)\) and \((2)\),X should be a perfect square and a palindrome.
X can be \(121(11^2), 484(22^2)\) as there is no unique value of X and unit digit of X can be even or odd, Combining (1) and (2) will also be insufficient.

Solution

Given:

• X is a 3-digit number

To find:

• Whether X is odd or not.

Analysing Statement 1
As per the information given in statement 1, X is a perfect square.

• Starting from 100 to 961, we have many even and odd perfect squares

Therefore, statement 1 is not sufficient

Analysing Statement 2
As per the information given in statement 2, X is palindrome.

• In 3-digits, we have many even as well as odd numbers which are palindromes.

Therefore, statement 2 is not sufficient.

Combining Both Statements
We know, X is a 3-digit perfect square, which is also a palindrome.
Possible values of X can be

• \(11^2 = 121\)
• \(22^2 = 484\)
• \(26^2 = 676\)

As we can see, X can be both even or odd.

Hence, the correct answer is option E.
_________________

1. X can be 121,144 etc. It can be even or odd. Hence insufficient
2. X can be 121,212 etc. It can be even or odd. Hence insufficient
Combining both statements X can be 121 or 484.
Hence answer is E

100≤X≤999

(1) X is a perfect square: X={121,144,169,196…}; insufic.
(2) X is a palindrome: X={121,222,313,505,808…}; insufic.
(1&2) X=perfect square-palindrome={11^2,(2*11)^2…}={121,484,…}; insufic.

Answer (E)